# Thread: two variabe induction question..

1. ## two variabe induction question..

n1 is the smallest whole number for which this inequality works :

(1+x)^n >1 +n*x + n*x^2

also i am given that x>0

find n1

and prove this inequality for every n=>n1 by induction.

the base case:

(1+x)^n1 >1+n1*x + n1*x^2

i think its correct because i was told that this inequality works for n1.

n=k step we presume that this equation is true :

equation 1: (1+x)^k >1 +k*x + k*x^2

n=k+1 step we need to prove this equation:
equation 2: (1+x)^(k+1) >1 +(k+1)*x + (k+1)*x^2

now i need to multiply equation1 by sum thing
and
do
if a<b<c
then a<c

how to do thing in this case?

2. equation 1: (1+x)^k >1 +k*x + k*x^2
n=k+1 step we need to prove this equation:
equation 2: (1+x)^(k+1) >1 +(k+1)*x + (k+1)*x^2

Just multiply equation 1 by (1+x) which is > 0
(1+x)^(k+1) >(1 +k*x + k*x^2) (1+x)
Then develop the second member

3. i tried to do this
but its not working
(1+x)^(k+1) >(1 +k*x + k*x^2) (1+x)>(1 +k*x + k*x^2)

(1+x)^(k+1) >(1 +k*x + k*x^2)

??

4. It is !
(1 + kx + kx²) (1+x) = 1 + (k+1)x + 2kx² + kx^3 > 1 + (k+1)x + 2kx² > 1 + (k+1)x + (k+1)x² because 2k > k+1

5. i need to prove that equation 2 is true:
(1+x)^(k+1) >1 +(k+1)*x + (k+1)*x^2

i take equation1 and multiply it by (1+x)

(2+x)^k >(1 +k*x + k*x^2)(1+x)

but i dont know how to involve equation2 in here

??

6. Originally Posted by transgalactic
i take equation1 and multiply it by (1+x)

(2+x)^k >(1 +k*x + k*x^2)(1+x)

but i dont know how to involve equation2 in here

??
Taking equation 1
(1+x)^k >1 +k*x + k*x^2
and multiplying by (1+x) gives
(1+x)^k (1+x)>(1 +k*x + k*x^2)(1+x)
(1+x)^(k+1)>1 + (k+1)x + 2kx² + kx^3 > 1 + (k+1)x + 2kx² > 1 + (k+1)x + (k+1)x² because 2k > k+1

Remember that (1+x)^k (1+x)=(1+x)^(k+1)

7. why (1+x)^k (1+x)=(1+x)^(k+1)

8. Because
(1+x)^k------=(1+x)(1+x) ... (1+x)--------------> k times
(1+x)^k (1+x)=(1+x)(1+x) ... (1+x)(1+x)---------> k+1 times

Therefore (1+x)^k (1+x)=(1+x)^(k+1)

9. aaahhh ok thanks

i am being asked to find n1

how to do that
??

10. Well you can see that it does not work for n=1 and n=2 but it works for n=3
I would say n1=3

11. i can replace n1 with n

(1+x)^n1 >1 +n1*x + n1*x^2

(1+x)^n1>1+n1*(x+x^2)

how to separate n1 here?

is this the right way??

12. I am afraid you cannot separate n1 in equation (1+x)^n1 >1 +n1*x + n1*x^2
The only way I can see to find n1 is the one I gave you in my previous post

13. i did that

if n=1
(1+x)>1+x+x^2

if n=2

(1+x)^2 >1 +2*x + 2*x^2

if n=3

(1+x)^3>1+3*x+3*x^2

these cases dont work

what to do?

14. I do not understand how you calculate

if n=1
(1+x)<1+x+x^2 => does not work

if n=2
(1+x)^2 = 1+2x+x² < 1 +2*x + 2*x^2 => does not work

if n=3
(1+x)^3 = 1+3x+3x²+x^3 >1+3*x+3*x^2 => works

15. thanks

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